Let $$\mathbf{a}$$ and $$\mathbf{b}$$ be two unit vectors such that $$|\mathbf{a} + \mathbf{b}| + 2|\mathbf{a} \times \mathbf{b}| = 2$$. If $$\theta \in (0, \pi)$$ is the angle between $$\hat{a}$$ and $$\hat{b}$$, then among the statements:
$$(S1) : 2|\hat{a} \times \hat{b}| = |\hat{a} - \hat{b}|$$
$$(S2)$$ : The projection of $$\hat{a}$$ on $$(\hat{a} + \hat{b})$$ is $$\frac{1}{2}$$

Let the angle between the unit vectors $$\hat a$$ and $$\hat b$$ be $$\theta$$.

Given,

$$\left|(\hat a+\hat b)+2(\hat a\times\hat b)\right|=2$$

Now,

$$\hat a+\hat b$$ lies in the plane containing $$\hat a,\hat b$$ while

$$\hat a\times\hat b$$ is perpendicular to that plane.

Hence,

$$ (\hat a+\hat b)\perp(\hat a\times\hat b) $$

Therefore, using

$$|\vec u+\vec v|^2=|\vec u|^2+|\vec v|^2$$

for perpendicular vectors,

$$\left|(\hat a+\hat b)+2(\hat a\times\hat b)\right|^2=|\hat a+\hat b|^2+|2(\hat a\times\hat b)|^2$$

Given modulus is $$2,$$ so

$$|\hat a+\hat b|^2+|2(\hat a\times\hat b)|^2=4$$

Now,

$$|\hat a+\hat b|^2=|\hat a|^2+|\hat b|^2+2\hat a\cdot\hat b$$

Since $$|\hat a|=|\hat b|=1,$$

$$|\hat a+\hat b|^2=1+1+2\cos\theta$$

Also,

$$|\hat a\times\hat b|=|\hat a||\hat b|\sin\theta=\sin\theta$$

Hence,

$$|2(\hat a\times\hat b)|^2=4\sin^2\theta$$

Substituting,

$$2+2\cos\theta+4\sin^2\theta=4$$

Using

$$\sin^2\theta=1-\cos^2\theta,$$

we get

$$2+2\cos\theta+4(1-\cos^2\theta)=4$$

$$2+2\cos\theta+4-4\cos^2\theta=4$$

$$4\cos^2\theta-2\cos\theta-2=0$$

Dividing by $$2,$$

$$2\cos^2\theta-\cos\theta-1=0$$

Factorising,

$$(2\cos\theta+1)(\cos\theta-1)=0$$

Thus,

$$\cos\theta=1\qquad \text{or}\qquad \cos\theta=-\frac12$$

Since the vectors are not parallel in the given condition, we take

$$\cos\theta=-\frac12$$

Hence,

$$\theta=\frac{2\pi}{3}=120^\circ$$

Now verify Statement $$(S1)$$:

$$2|\hat a\times\hat b|=|\hat a-\hat b|$$

LHS:

$$2|\hat a\times\hat b|=2\sin120^\circ$$

$$=2\cdot\frac{\sqrt3}{2}=\sqrt3$$

RHS:

$$|\hat a-\hat b|=\sqrt{|\hat a|^2+|\hat b|^2-2\hat a\cdot\hat b}$$

$$=\sqrt{1+1-2\cos120^\circ}$$

$$=\sqrt{2-2\left(-\frac12\right)}$$

$$=\sqrt3$$

Thus,

$$2|\hat a\times\hat b|=|\hat a-\hat b|$$

Hence, $$(S1)$$ is true.

Now verify Statement $$(S2)$$:

The projection of $$\hat a$$ on $$\hat a+\hat b$$ is

$$\frac{\hat a\cdot(\hat a+\hat b)}{|\hat a+\hat b|}$$

$$=\frac{\hat a\cdot\hat a+\hat a\cdot\hat b}{\sqrt{|\hat a|^2+|\hat b|^2+2\hat a\cdot\hat b}}$$

$$=\frac{1+\cos120^\circ}{\sqrt{2+2\cos120^\circ}}$$

$$=\frac{1-\frac12}{\sqrt{2-1}}$$

$$=\frac12$$

Hence, $$(S2)$$ is also true.

Therefore, both statements are true.

Hence, the correct answer is $$\boxed{\text{Both (S1) and (S2) are true}}$$.